Thread: Slope of the tangent line

1. Slope of the tangent line

The slope of the tangent live to a curve is given by f(x)=3x^2+5 if the point
(1,2) is on the curve, find the equation of the curve.
if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
So how would i go about finding the constant ?

2. Originally Posted by xterminal01
The slope of the tangent live to a curve is given by f(x)=3x^2+5 if the point
(1,2) is on the curve, find the equation of the curve.
if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
So how would i go about finding the constant ?
plug in the initial condition ...

$\displaystyle f(x) = x^3 + 5x + C$

$\displaystyle f(1) = 1^3 + 5(1) + C$

$\displaystyle 2 = 6 + C$

$\displaystyle C = -4$

3. Originally Posted by xterminal01
The slope of the tangent live to a curve is given by f`(x)=3x^2+5 if the point
(1,2) is on the curve, find the equation of the curve.
if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
So how would i go about finding the constant ?
(1,2) <---this is your initial condition

$\displaystyle (1,2)\text{ on f(x) }\Rightarrow2=(1)^3+5(1)+C$